Question

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample.

Short answer

a. The 90% confidence interval for the proportion of adult Americans who thought lying was never justified can be estimated using the formula and is between two values. b. The same process gives the 90% confidence interval for the proportion of adult Americans who thought that lying to not hurt feelings was often or sometimes OK, again between two values. c. The overlap or lack thereof in these intervals will indicate whether the majority of adult Americans have consistent or inconsistent stances on the different types of lying.

Step-by-step solution

- Construct a 90% confidence interval for the proportion of adult Americans who think lying is never justified

Firstly, calculate the sample proportion (\(p̂\)) where \(p̂ = \frac{x}{n}\). Here, \(x\) is the number of successes (people who think lying is never justified) and \(n\) is the total number of observations. For this part of the problem, \(x = 0.52 * 1000 = 520\) and \(n = 1000\), hence \(p̂ = \frac{520}{1000} = 0.52\). Then, calculate the Z-score for a 90% confidence interval (\(Z_{\frac{\alpha}{2}}\) for 90% CI is 1.645). Now using the formula for a confidence interval, we calculate \(CI = 0.52 \pm 1.645 \sqrt{\frac{0.52(1-0.52)}{1000}}\).

- Construct a 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings

Again, calculate the sample proportion (\(p̂\)) where \(p̂ = \frac{x}{n}\). For this part of the problem, \(x = 650\) and \(n = 1000\), hence \(p̂ = \frac{650}{1000} = 0.65\). The Z-score for a 90% confidence interval remains the same (\(Z_{\frac{\alpha}{2}} = 1.645\)). Now using the confidence interval formula, we calculate \(CI = 0.65 \pm 1.645 \sqrt{\frac{0.65(1-0.65)}{1000}}\).

- Comment on the apparent inconsistency in the responses

Now, compare the two confidence intervals. If they overlap, the results can still be consistent despite seeming contradictory. If they do not overlap, it indicates an inconsistency in the responses. Remember that a confidence interval reflects the range in which we expect the true population parameter to exist given the observed data and the level of confidence.